Answer:
The possible zeroes are -3 , -5/3 and 1/2
Step-by-step explanation:
* To find the zeroes you must let f(x) = 0
* First step look at the numerical term -15 then find two numbers multiply by each other and = -15⇒[ (5×-3) (-5×3) (-1×15) (1×-15)] and let x = one of them
* Second step Chose on of them and substitute x by it, I will chose -3
* f(-3) = [tex]6(-3^{3})+25(-3^{2})+16(-3)-15=-162+225-48-15=0[/tex]
∵ f(x) = 0 when x = -3
∴ x + 3 is a factor of f(x)
* Third step divide f(x) by its factor to get quadratic and factorize it
[tex]\frac{6x^{3}+25x^{2}+16x-15 }{x+3}=6x^{2}\frac{7x^{2}+16x-15 }{x+3}[/tex]
[tex]\frac{7x^{2}+16x-15 }{x+3}=7x\frac{-5x-15}{x+3}[/tex]
[tex]\frac{-5x-15}{x+3}=-5[/tex]
∴ f(x) = (x + 3)([tex]6x^{2}+7x-5[/tex]⇒factorize the quadratic
∵f(x) = (x + 3)(2x - 1)(3x + 5)
∵ f(x) = 0
∴ x + 3 = 0 ⇒ x = -3
∴ 2x - 1 = 0 ⇒ x = 1/2
∴ 3x + 5 = 0 ⇒ x = -5/3
∴
